Problem 293: Pseudo-Fortunate Numbers

An even positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes.
The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48.

If N is admissible, the smallest integer M > 1 such that N+M is prime, will be called the pseudo-Fortunate number for N.

For example, N=630 is admissible since it is even and its distinct prime factors are the consecutive primes 2,3,5 and 7.
The next prime number after 631 is 641; hence, the pseudo-Fortunate number for 630 is M=11.
It can also be seen that the pseudo-Fortunate number for 16 is 3.

Find the sum of all distinct pseudo-Fortunate numbers for admissible numbers N less than 10^9.

My Algorithm

The problem description starts with "an even positive integer [...]" and continues with "[...] its distinct prime factors are consecutive primes.".
Therefore any admissible number must be a multiple of 2 and its prime factors must start with 2, 3, 5, 7, ... without any gap.
Its general form is: 2^a * 3^b * 5^c * ... where a exponent is only zero if all following exponents are zero, too.
The highest relevant base (prime) is 23 because 2^1 * 3^1 * 5^1 * 7^1 * 9^1 * 11^1 * 13^1 * 17^1 * 19^1 * 23^1 = 223092870 < 10^9.
Extending the series with 29^1 would exceed 10^9.

In step 1 of my algorithm I create all such admissible numbers up to 10^9 in ascending order:

put a 2 in the automatically sorted todo container, its highest prime factor is 2 (index 0 of factors)

then repeat until todo is empty:

pick the lowest number from todo and store it in admissible

multiply the same number by all values of factor[0 ... index+1] (those are the prime factors) and store them in todo (if < 10^9)

In step 2 I run a simple prime sieve (trial division) to find all prime numbers up to sqrt{10^9}.
Step 3 needs these prime numbers to run a primality test on each admissible number's right neighbors.
Admissible numbers are always even, therefore only the numbers +3, +5, +7, +9, ... have to be checked.

Note

There are a few optimization tricks:

some admissible numbers are smaller than the prime numbers generated in step 2 → simple linear search in primes

for this specific range, not all primes in step 2 are actually needed → it's sufficient to have all primes up to sqrt{2.8 * 10^8} → 30% faster

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent toecho 50 | ./293

Output:

(please click 'Go !')

Note: the original problem's input 1000000000cannot be enteredbecause just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include<iostream>

#include<vector>

#include<map>

#include<set>

intmain()

{

unsignedint limit = 1000000000;

std::cin>> limit;

// ---------- step 1: find all admissible numbers ----------

// generate all numbers 2^a * 3^b * 5^c * 7^d * 11^e ...

constunsignedint NumFactors = 9;

constunsignedint factors[NumFactors] = { 2,3,5,7,11,13,17,19,23 };

// product of all factors is just under 10^9, including the next prime (29) would be bigger than 10^9

// store all admissible numbers

std::vector<unsignedint> admissible;

// [number] => [index of its highest prime factor]

std::map<unsignedint, unsignedchar> todo = { { 2, 0 } };

while (!todo.empty())

{

// pick smallest number and remove from to-do list

auto current = *(todo.begin());

todo.erase(todo.begin());

auto number = current.first;

auto maxPrime = current.second;

// add to admissible list

admissible.push_back(number);

// multiply by all already used prime numbers and the next one

for (unsignedchar i = 0; i <= maxPrime + 1; i++) // can start with i = maxPrime, too

Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.

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[new]

the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.

The 310 solved problems (that's level 12) had an average difficulty of 32.6&percnt; at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of &approx;60000 in August 2017)
at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.All of my solutions can be used for any purpose and I am in no way liable for any damages caused.You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.Thanks for all their endless effort !!!

more about me can be found on my homepage,
especially in my coding blog.
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